Let assume that the mean is 184 and the standard deviation is 6
Heights of men on a baseball team have a bell-shaped distribution with a mean 184 of and a standard deviation of 6 . Using the empirical rule, what is the approximate percentage of the men between the following values? a.166 cm and 202 cm b. 172 cm and 196cm
Answer:
P(156<X<202) = 99.7%
P(172<X<196) = 95.5%
Step-by-step explanation:
Given that :
Heights of men on a baseball team have a bell-shaped distribution with a mean of and a standard deviation of . Using the empirical rule, what is the approximate percentage of the men between the following values? a.166 cm and 202 cm b. 172 cm and 196cm
For a.
Using the empirical rule, what is the approximate percentage of the men between the following values 166 cm and 202 cm.
the z score can be determined by using the formula:
[tex]z = \dfrac{X - \mu}{\sigma}[/tex]
[tex]z(166) = \dfrac{166-184}{6}[/tex]
[tex]z(166) = \dfrac{-18}{6}[/tex]
z(166) = -3
[tex]z(202) = \dfrac{202-184}{6}[/tex]
[tex]z(202) = \dfrac{18}{6}[/tex]
z(202) = 3
P(156<X<202) = P( μ - 3σ < X < μ + 3σ )
P(156<X<202) = P( - 3 < Z < 3)
P(156<X<202) = P( Z < 3) - P(Z < -3)
P(156<X<202) = 0.99865- 0.001349
P(156<X<202) = 0.997301
P(156<X<202) = 99.7%
For b.
b. 172 cm and 196cm
[tex]z = \dfrac{X - \mu}{\sigma}[/tex]
[tex]z(172) = \dfrac{172-184}{6}[/tex]
[tex]z(172) = \dfrac{-12}{6}[/tex]
z(172) = -2
[tex]z(196) = \dfrac{196-184}{6}[/tex]
[tex]z(196) = \dfrac{12}{6}[/tex]
z(196) = 2
P(172<X<196) = P( μ - 2σ < X < μ + 2σ )
P(172<X<196) = P( - 2 < Z < 2)
P(172<X<196) = P( Z < 2) - P(Z < -2)
P(172<X<196) = 0.9772 - 0.02275
P(172<X<196) = 0.95445
P(172<X<196) = 95.5%
